discrete time signal(particular and homogenous solution

Thread Starter

TAKYMOUNIR

Joined Jun 23, 2008
352
can someone help me to know how to find particular solution and homogenous solution for discrete time signal ,if you know links for that subject ,i did search on google but i can not find something
thanks
 

blah2222

Joined May 3, 2010
582
Are you referring to difference equations, which are analogous to differential equations for continuous-time signals?

Tons of stuff if you know how to use Google.

Is there a particular example that you could work through and post on here? That might be more useful.

JP
 

Thread Starter

TAKYMOUNIR

Joined Jun 23, 2008
352
like y[n]=y[n-1]+y[n-2]+x[n]
where x[n]=n^2*u[n]
u[n] is unit step function

Are you referring to difference equations, which are analogous to differential equations for continuous-time signals?

Tons of stuff if you know how to use Google.

Is there a particular example that you could work through and post on here? That might be more useful.

JP
 

blah2222

Joined May 3, 2010
582
like y[n]=y[n-1]+y[n-2]+x[n]
where x[n]=n^2*u[n]
u[n] is unit step function
Okay, are you given any initial conditions of y[n] for n < 0? If not, assume y[-2] = 0 and y[-1] = 0 then go through each value of y[n] for n.

y[0] = y[-1] + y[-2] + x[0] = 0 + 0 + 0^2*u[0] = 0 + 0 + 0*1 = 0
y[1] = ...

Do this until the values converge and stop changing.

In the example that you have given, if there are no initial conditions your function will be a zero function for all n.
 

Thread Starter

TAKYMOUNIR

Joined Jun 23, 2008
352
y[n]=yP[n]+yh[n]
where yp[n] is the particular solution and yh[n] is the homogenous solution so i find this answer yp and yh
yh=k^n
yp=m*u[n]
so why this answer

Okay, are you given any initial conditions of y[n] for n < 0? If not, assume y[-2] = 0 and y[-1] = 0 then go through each value of y[n] for n.

y[0] = y[-1] + y[-2] + x[0] = 0 + 0 + 0^2*u[0] = 0 + 0 + 0*1 = 0
y[1] = ...

Do this until the values converge and stop changing.

In the example that you have given, if there are no initial conditions your function will be a zero function for all n.
 
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